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In my study of ODEs I have recently encountered this monster of an integral the sum of two integrals: $ \int_{0}^{2\pi} \frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx + \int_{0}^{2\pi} \frac{\sin(x)\cos\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx $

Even Wolfram is having trouble with this, I was hoping someone could please tell me if I can do it analytically or numerically any way at all? It really is scary looking. Of course $\omega$ and $A$ are parameters with $A$ between 0 and 1.

Edits:

The first one (with the cosine) is zero as the great answers point out but the one with the product of two sines is a mystery.

I feel I owe an explanation: the relation to ODE comes from a previous problem I stated here dealing with the Melnikov integral here the answer changes variables from t to x.

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    $\begingroup$ You might find a nice substitution to simplify the argument of the arctanh. Maybe $tanh(u) = $ everything inside the arctanh? That might make things worse though. $\endgroup$
    – Paul Castle
    Commented Jul 11, 2016 at 7:07
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    $\begingroup$ Um... What exactly are you trying to do here with ODE's? Does this integral arise from solving a differential equation? If so, it might help to see the original problem. $\endgroup$
    – Vegeta the Prince of Saiyans
    Commented Jul 11, 2016 at 7:38
  • $\begingroup$ @LordVader007 : This integral is not directly from an ODE rather the Melnikov integral relating to a perturbation of an ODE what matters is the improper integral $\endgroup$
    – Croc2Alpha
    Commented Jul 11, 2016 at 8:34
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    $\begingroup$ It is really bad style to edit an answered question into a basically completely different one. You should have asked a new question for that $\endgroup$
    – Tobias Kienzler
    Commented Jul 11, 2016 at 11:33

2 Answers 2

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It's simpler than it appears at first :

Let $\quad f(x)=\frac{\sin(x)\cos\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}$

$$\int_0^{2\pi} f(x)dx = \int_0^{\pi} f(x)dx +\int_{\pi}^{2\pi} f(x)dx $$ Let $\quad x=2\pi-t \quad$ It is easy to prove that $\quad f(x)=-f(t)$

because $\sin(x)=\sin(-t)=-\sin(t)$ and $\sin^2\left(\frac{x}{2}\right)=\sin^2\left(\pi-\frac{t}{2}\right)=\sin^2\left(\frac{t}{2}\right)$

$\int_{\pi}^{2\pi} f(x)dx =\int_{\pi}^{0} \left(-f(t)\right)(-dt) =-\int_{0}^{\pi} f(t)dt $ $$\int_0^{2\pi} f(x)dx = \int_0^{\pi} f(x)dx -\int_{0}^{\pi} f(t)dt=0 $$

$$\int_0^{2\pi}\frac{\sin(x)\cos\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx=0$$

With the same method and $g(x)=\frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx$

$g(x)=g(t) \quad\to\quad \int_0^{2\pi}g(x)dx=\int_0^{\pi}g(x)dx+\int_0^{\pi}g(t)dt=2\int_0^{\pi}g(x)dx\neq 0$

$$\int_0^{2\pi}\frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx \neq 0$$

In the equation above, the symbol $\neq 0$ means "generally not equal to $0$ ". It doesn't mean "always not equal to $0$ ".

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    $\begingroup$ I cannot understand "what about the -". what"-" ? $\endgroup$
    – JJacquelin
    Commented Jul 11, 2016 at 8:59
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    $\begingroup$ Thank you for a brilliant answer, I wanted to ask in the comment as I broke up $ \sin{(\omega(t+t_0))} $ via a trig identity how would I handle the integral same as above only with $ \sin{(\omega t(x))} $ ? I mean integrating on the same interval $ \quad f(x)=\frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}$ $\endgroup$
    – Croc2Alpha
    Commented Jul 11, 2016 at 9:04
  • $\begingroup$ is it the same method for what I mentioned as well? $\endgroup$
    – Croc2Alpha
    Commented Jul 11, 2016 at 9:09
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    $\begingroup$ Can you write down clearly the whole integral you want to solve, with $dt$ or $dx$ in order to avoid confusion. $\endgroup$
    – JJacquelin
    Commented Jul 11, 2016 at 9:24
  • $\begingroup$ sorry about the mess here is the second one I forgot about $\int_{0}^{2\pi} \frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx$ $\endgroup$
    – Croc2Alpha
    Commented Jul 11, 2016 at 9:30
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Oh, the answer is 0. Note that the function satisfies $$f(x) = - f(2\pi - x)$$ for any values of $A, \Omega$.

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    $\begingroup$ Thank you could you please expand on how you got this? $\endgroup$
    – Croc2Alpha
    Commented Jul 11, 2016 at 8:54
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    $\begingroup$ I tried graphing the curve with some example values, and I noticed the nice symmetry around $\pi$. $\endgroup$
    – Paul Castle
    Commented Jul 11, 2016 at 8:59
  • $\begingroup$ Thanks could you please help with the next part integrating the function $ \quad f(x)=\frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{‌​2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} $ on the same interval $\endgroup$
    – Croc2Alpha
    Commented Jul 11, 2016 at 9:15
  • $\begingroup$ The $f(x)$ is different from the preceding one. This might cause another mess, so it should be better to use a different symbol : $g(x)=\int_0^{2\pi}\frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx$. $$\int_0^{2\pi}f(x)dx=0 \text{ and } \int_0^{2\pi}g(x)dx \neq 0$$ $\endgroup$
    – JJacquelin
    Commented Jul 11, 2016 at 10:24
  • $\begingroup$ The function $\frac{\sin x}{\sqrt{A \sin^2 \left(\frac{x}{2} \right)}}$ has a relatively simple integral. It's possible that an integration by parts will let you make some kind of symmetry argument for $g(x)$ like the one we made for $f(x)$. I don't have time to try it out though. $\endgroup$
    – Paul Castle
    Commented Jul 12, 2016 at 0:44

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